If so, what would be the most efficient algorithm for generating spheres with different number of hexagonal faces at whatever interval required to make them fit uniformly or how might you calculate how many hexagonal faces are required for each subdivision?
|
|
No, not even if we permit non-regular hexagonal faces. The reason is more graph-theoretical than geometrical. We begin with Euler's formula, relating the number of faces $F$, the number of vertices $V$, and the number of edges $E$: $$ F+V-E = 2 $$ Consider the faces meeting at a vertex. This number must be at least three for any vertex (two faces cannot meet at a vertex in a solid). Thus, if we add up the six vertices for each hexagonal face, we will count each vertex at least three times. That is to say, $$ V \leq \frac{6F}{3} = 2F $$ On the other hand, if we add up the six edges for each hexagonal face, we will count each edge exactly twice, so that $$ E = \frac{6F}{2} = 3F $$ Substituting these into Euler's formula, we obtain $$ F+V-E \leq F+2F-3F = 0 $$ But if $F+V-E \leq 0$, then it is impossible that $F+V-E = 2$, so no solid can be composed solely of hexagonal faces, even if we permit non-regular hexagons. If we now restrict ourselves to regular faces, we can show an interesting fact: Any solid with faces made up of nothing other than regular hexagons and pentagons must have exactly $12$ pentagons on it (the limiting case being the hexagon-free dodecahedron). Again, we begin with Euler's formula: $$ F+V-E = 2 $$ Let $F_5$ be the number of pentagonal faces, and $F_6$ be the number of hexagonal faces. Then $$ F = F_5+F_6 $$ The only number of faces that can meet at a vertex is $3$; there isn't enough angular room for $4$ faces to meet, and solids can't have $2$ faces meeting at a vertex. If we add up the five vertices of each pentagon and the six vertices of each hexagon, then we have counted each vertex three times. That is to say, $$ V = \frac{5F_5+6F_6}{3} $$ Similarly, if we count up the five edges of each pentagon and the six edges of each hexagon, then we have counted each edge twice, and so, $$ E = \frac{5F_5+6F_6}{2} $$ Plugging these expressions back into Euler's formula, we obtain $$ F_5+F_6+\frac{5F_5+6F_6}{3}-\frac{5F_5+6F_6}{2} = 2 $$ The $F_6$ terms cancel out, leaving $$ \frac{F_5}{6} = 2 $$ or just $F_5 = 12$. I've heard tell that any number of hexagonal faces $F_6$ is permitted except $F_6 = 1$, but I haven't confirmed that for myself. The basic line of reasoning for excluding $F_6 = 1$ may be as follows: Suppose a thirteen-sided polyhedron with one hexagonal face and twelve pentagonal faces exists. Consider the hexagonal face. It must be surrounded by six pentagonal faces; call these $A$ through $F$. Those pentagonal faces describe, at their "outer" edge, a perimeter with twelve edges and twelve vertices, which must be shared by a further layer of six pentagonal faces; call these $G$ through $L$. There cannot be fewer than this, because the twelve edges are arranged in a cycle of six successive pairs, each pair belonging to one of $A$ through $F$. No two faces can share more than one edge, so the twelve edges must be shared amongst six faces $G$ through $L$, but "out of phase" with $A$ through $F$. However, these pentagonal faces $G$ through $L$ cannot terminate in a single vertex—they would have to be squares to do that. Hence, they must terminate in a second hexagon. Thus, a polyhedron of the type envisioned cannot exist. Likely the above approach could be made more rigorous, or perhaps there is a more clever demonstration. |
|||||||||||||||||||||
|
|
You can do it if you allow your sphere to have a hole through it, i.e. be a torus. See image here. Then the Euler characteristic, as described in Brian Tung's excellent answer is 0, not 2. You might also be able to do it for non-convex polygons; convexity is a requirement of the Euler characteristic. See, e.g., the octahemioctahedron. I couldn't find one with just hexagons though. |
|||
|
|
